Problem: Divide the following complex numbers: $\dfrac{6(\cos(\frac{23}{12}\pi) + i \sin(\frac{23}{12}\pi))}{3(\cos(\frac{19}{12}\pi) + i \sin(\frac{19}{12}\pi))}$ (The dividend is plotted in blue and the divisor in plotted in green. Your current answer will be plotted orange.)
Solution: Dividing complex numbers in polar forms can be done by dividing the radii and subtracting the angles. The first number ( $6(\cos(\frac{23}{12}\pi) + i \sin(\frac{23}{12}\pi))$ ) has angle $\frac{23}{12}\pi$ and radius 6. The second number ( $3(\cos(\frac{19}{12}\pi) + i \sin(\frac{19}{12}\pi))$ ) has angle $\frac{19}{12}\pi$ and radius 3. The radius of the result will be $\frac{6}{3}$ , which is 2. The angle of the result is $\frac{23}{12}\pi - \frac{19}{12}\pi = \frac{1}{3}\pi$ The radius of the result is $2$ and the angle of the result is $\frac{1}{3}\pi$.